Re: prop rpm question

[Bob Fry]

Uh, thanks for trying...I guess:

DC> "Bob Fry" > wrote in message
DC> ...
>> At 1000 rpm or so, my airplane will taxi and get up to, what,
>> 15-20 kts? But at double the rpm it will fly at 80-90 kts,
>> though it would take a long time to take off. Surely double
>> the rpm produces more than double the propellor thrust...or
>> does it? Anyway, it seems very nonlinear, that is, double the
>> rpm and I get much more than double the performance. Why is
>> that?

DC> The laws of physics (certainly those relating to mechanics -
DC> velocity, acceleration, thrust, drag and all that) are rarely
DC> linear.

Eh? F=ma and many others are.

DC> On the ground your aircraft is probably not in an
DC> optimum attitude for drag reduction, so it's probably not fair
DC> to compare it with an aircraft in the sky. And will your
DC> aircraft fly straight and level at 1000rpm? If so, how fast?

Kee-rist. Drag, and attitude (angle of attack, really) have little to
do with the explanation I was looking for. No, of course it won't fly
straight and level at 1000 rpm. That's nearly full idle landing rpm.

DC> The main thing dictating how your aircraft performs is
DC> drag. "Normal" drag increases with the square of the speed you
DC> fly at

I think you mean parasitic drag.

DC> - so if you double the speed, you roughly quadruple the
DC> drag (hence everything has a terminal velocity when falling to
DC> earth - as you get faster, the drag increases faster than your
DC> speed increases, and you stop accelerating once drag equals
DC> the acceleration caused by gravity). Remember also that at low
DC> speeds you have induced drag, which is high at low speeds but
DC> vanishes as you get faster.

Induced drag--drag caused by the wing producing lift at a vector not
perpendicular to flight--never vanishes unless lift vanishes.

Anyway I'll restate the question, plus post to r.a.'s garbage heap,
r.a.piloting.

At 1000 rpm the prop produces some amount of thrust (lift), call it
T[1000]. This thrust is only enough to move the plane in a moderate
taxi.

At double that rpm, 2000 rpm, the prop produces another amount of
thrust, call it T[2000]. Now I'm not positive, but it sure seems that

T[2000] >> T[1000]

Why, if rpm only doubles, does thrust (seem to) much more than double?

[Kyle Boatright]

"Bob Fry" > wrote in message
...
> Uh, thanks for trying...I guess:
>>>snip>>>
> Anyway I'll restate the question, plus post to r.a.'s garbage heap,
> r.a.piloting.
>
> At 1000 rpm the prop produces some amount of thrust (lift), call it
> T[1000]. This thrust is only enough to move the plane in a moderate
> taxi.
>
> At double that rpm, 2000 rpm, the prop produces another amount of
> thrust, call it T[2000]. Now I'm not positive, but it sure seems that
>
> T[2000] >> T[1000]
>
> Why, if rpm only doubles, does thrust (seem to) much more than double?

The short way around is to look at your engine's power chart and evaluate
how many excess HP it is developing at 1000 rpm vs 2,000 rpm.

If we assume the plane in question is a C-152, the engine is making very
little power at 1,000 rpm. I'd guess 15 hp, of which at least 5 hp is spent
in friction inside the engine, leaving 10 hp for thrust. At 2,000 rpm, the
engine is probably making 60 hp, of which 10 is spent on internal friction.
Therefore, you have 50 hp for thrust.

Another way to look at it is that your prop has an advance rate. Let's say
it the advance rate is 4 feet per revolution. At 1,000 rpm, and no drag on
the airplane (rolling or aerodynamic), the airplane would have a terminal
velocity of 4,000 fpm, or about 48 mph. Of course, there is rolling and
aerodynamic drag, and there is prop drag too, so the engine can only drag
the plane along at, say, 30 mph, assuming a flat smooth runway.

At 2,000 rpm, with no drag, the terminal velocity would be 8,000 fpm, or
about 85 mph. Of course, there is still aerodynamic and prop drag, but there
is no rolling resistance, so you get more bang for your RPM buck. Of course,
it helps that your engine is delivering 60 hp, as opposed to 15 hp when it
turns 1,000 rpm.

Yet another way to look at it is that when your prop spins at 2x the speed,
it requires 4x the power to turn it... KE=1/2MV^2..

And yes, all of this stuff is ideal world, no prop efficiency losses, etc...

KB

KB

[Bob Fry]

I wish to leave the engine out of the discussion, but let's
continue...

>>>>> "KB" == Kyle Boatright > writes:

KB> If we assume the plane in question is a C-152,

Close enough, it's an Aircoupe with a C90.

But let's look just at the prop. Why does a prop produce so much more
thrust, much more than double, when it's turned at only twice the
rate?

KB> Another way to look at it is that your prop has an advance
KB> rate. Let's say it the advance rate is 4 feet per
KB> revolution.

Yep, 48" pitch.

KB> At 1,000 rpm, and no drag on the airplane (rolling
KB> or aerodynamic), the airplane would have a terminal velocity
KB> of 4,000 fpm, or about 48 mph. Of course, there is rolling and
KB> aerodynamic drag, and there is prop drag too, so the engine
KB> can only drag the plane along at, say, 30 mph, assuming a flat
KB> smooth runway.

KB> At 2,000 rpm, with no drag, the terminal velocity would be
KB> 8,000 fpm, or about 85 mph.

Hmmmm...so prop thrust is indeed only twice at double the
rpm?...ideally speaking of course.

The idealized (no viscosity etc.) math seems to say that it is linear,
but intuitive feel says not.

[skyloon]

This won't be a complete ( or maybe even correct answer) but ...

The prop is an airfoil. Here *lift* will be *thrust*. The thrust force
generated will be

T = C A d/2 V^2. V is the speed of the airfoil (prop). So the force is
nonlinear, it goes as the square.
For the rest of the parameters: C is coeff of lift (depends on angle of
attack), A is airfoil area, d= air density.

The prop will have induced and parasitic drag. The relative wind angle of
attack on the prop foil will depend on the aircraft speed. At low speed it
is at high angle of attack. The induced drag is large and the engine can't
get up to full rpm. As the aircraft speed increases, the AOA decreases, the
induced drag decreases and the can rev up to higher rpm (as constnt
throttle). There will be an AOA where the drag is minimum, This is where the
prop is most efficient. It is a narrow range, because, as you go faster,
parasitic drag on the prop kicks in. Constant speed props are used to adjust
the pitch to remian efficient over a wider raneg of airspeeds.

Kershner states that the maximum thrust force occurs when the plane is
standing still (at a fixed throttle setting, I guess), and decreases as you
go faster. I do not understand this. Is it beacese AOA is largest? I am
trying to see how this relates to power. Power would be force*distance/time
or force*velocity. Maybe the thrust decreases slowly with airspeed, but the
power still goes up as you go faster.

This is just a hand waving argument. Please, anyone who knows more, feel
free to correct this picture.

Dave






"Bob Fry" > wrote in message
...
>I wish to leave the engine out of the discussion, but let's
> continue...
>
>>>>>> "KB" == Kyle Boatright > writes:
>
> KB> If we assume the plane in question is a C-152,
>
> Close enough, it's an Aircoupe with a C90.
>
> But let's look just at the prop. Why does a prop produce so much more
> thrust, much more than double, when it's turned at only twice the
> rate?
>
> KB> Another way to look at it is that your prop has an advance
> KB> rate. Let's say it the advance rate is 4 feet per
> KB> revolution.
>
> Yep, 48" pitch.
>
> KB> At 1,000 rpm, and no drag on the airplane (rolling
> KB> or aerodynamic), the airplane would have a terminal velocity
> KB> of 4,000 fpm, or about 48 mph. Of course, there is rolling and
> KB> aerodynamic drag, and there is prop drag too, so the engine
> KB> can only drag the plane along at, say, 30 mph, assuming a flat
> KB> smooth runway.
>
> KB> At 2,000 rpm, with no drag, the terminal velocity would be
> KB> 8,000 fpm, or about 85 mph.
>
> Hmmmm...so prop thrust is indeed only twice at double the
> rpm?...ideally speaking of course.
>
> The idealized (no viscosity etc.) math seems to say that it is linear,
> but intuitive feel says not.
>

[Kyle Boatright]

"Bob Fry" > wrote in message
...
>I wish to leave the engine out of the discussion, but let's
> continue...
>
>>>>>> "KB" == Kyle Boatright > writes:
>
> KB> If we assume the plane in question is a C-152,
>
> Close enough, it's an Aircoupe with a C90.
>
> But let's look just at the prop. Why does a prop produce so much more
> thrust, much more than double, when it's turned at only twice the
> rate?
>
> KB> Another way to look at it is that your prop has an advance
> KB> rate. Let's say it the advance rate is 4 feet per
> KB> revolution.
>
> Yep, 48" pitch.
>
> KB> At 1,000 rpm, and no drag on the airplane (rolling
> KB> or aerodynamic), the airplane would have a terminal velocity
> KB> of 4,000 fpm, or about 48 mph. Of course, there is rolling and
> KB> aerodynamic drag, and there is prop drag too, so the engine
> KB> can only drag the plane along at, say, 30 mph, assuming a flat
> KB> smooth runway.
>
> KB> At 2,000 rpm, with no drag, the terminal velocity would be
> KB> 8,000 fpm, or about 85 mph.
>
> Hmmmm...so prop thrust is indeed only twice at double the
> rpm?...ideally speaking of course.

The thrust is probably 4x, like the engine's power. The advance rate makes
the prop want to pull the airplane at 2x the speed, but again, 2x the speed
requires 4x the power...



> The idealized (no viscosity etc.) math seems to say that it is linear,
> but intuitive feel says not.
>

[[email protected]]

: The thrust is probably 4x, like the engine's power. The advance rate makes
: the prop want to pull the airplane at 2x the speed, but again, 2x the speed
: requires 4x the power...

Check your equations. Drag force goes with the square of the velocity.
Drag power (force*velocity) therefore goes with the *cube* of velocity. So, if that's
the governing equation, 2x the speed requires 8x the power.

If you look at the operator handbooks for identical airframes with different
engine options (e.g. PA28-140/150/160/180/235, PA24-180/250/260/400), you'll see that
they almost exactly follow this cubic (i.e. cube-root) equation.

-Cory


--

************************************************** ***********************
* Cory Papenfuss *
* Electrical Engineering candidate Ph.D. graduate student *
* Virginia Polytechnic Institute and State University *
************************************************** ***********************

[Matt Whiting]

Bob Fry wrote:

> Uh, thanks for trying...I guess:
>
> DC> "Bob Fry" > wrote in message
> DC> ...
> >> At 1000 rpm or so, my airplane will taxi and get up to, what,
> >> 15-20 kts? But at double the rpm it will fly at 80-90 kts,
> >> though it would take a long time to take off. Surely double
> >> the rpm produces more than double the propellor thrust...or
> >> does it? Anyway, it seems very nonlinear, that is, double the
> >> rpm and I get much more than double the performance. Why is
> >> that?
>
> DC> The laws of physics (certainly those relating to mechanics -
> DC> velocity, acceleration, thrust, drag and all that) are rarely
> DC> linear.
>
> Eh? F=ma and many others are.
>
> DC> On the ground your aircraft is probably not in an
> DC> optimum attitude for drag reduction, so it's probably not fair
> DC> to compare it with an aircraft in the sky. And will your
> DC> aircraft fly straight and level at 1000rpm? If so, how fast?
>
> Kee-rist. Drag, and attitude (angle of attack, really) have little to
> do with the explanation I was looking for. No, of course it won't fly
> straight and level at 1000 rpm. That's nearly full idle landing rpm.
>
> DC> The main thing dictating how your aircraft performs is
> DC> drag. "Normal" drag increases with the square of the speed you
> DC> fly at
>
> I think you mean parasitic drag.
>
> DC> - so if you double the speed, you roughly quadruple the
> DC> drag (hence everything has a terminal velocity when falling to
> DC> earth - as you get faster, the drag increases faster than your
> DC> speed increases, and you stop accelerating once drag equals
> DC> the acceleration caused by gravity). Remember also that at low
> DC> speeds you have induced drag, which is high at low speeds but
> DC> vanishes as you get faster.
>
> Induced drag--drag caused by the wing producing lift at a vector not
> perpendicular to flight--never vanishes unless lift vanishes.
>
> Anyway I'll restate the question, plus post to r.a.'s garbage heap,
> r.a.piloting.
>
> At 1000 rpm the prop produces some amount of thrust (lift), call it
> T[1000]. This thrust is only enough to move the plane in a moderate
> taxi.
>
> At double that rpm, 2000 rpm, the prop produces another amount of
> thrust, call it T[2000]. Now I'm not positive, but it sure seems that
>
> T[2000] >> T[1000]
>
> Why, if rpm only doubles, does thrust (seem to) much more than double?

Because prop thrust increases as the square of the RPM and thus you get
four times more thrust at twice the RPM. There are other factors that
come into play as well such as advance ratio, but I assume you can Google...

Matt

[Matt Whiting]

Bob Fry wrote:

> I wish to leave the engine out of the discussion, but let's
> continue...
>
>
>>>>>>"KB" == Kyle Boatright > writes:
>
>
> KB> If we assume the plane in question is a C-152,
>
> Close enough, it's an Aircoupe with a C90.
>
> But let's look just at the prop. Why does a prop produce so much more
> thrust, much more than double, when it's turned at only twice the
> rate?
>
> KB> Another way to look at it is that your prop has an advance
> KB> rate. Let's say it the advance rate is 4 feet per
> KB> revolution.
>
> Yep, 48" pitch.
>
> KB> At 1,000 rpm, and no drag on the airplane (rolling
> KB> or aerodynamic), the airplane would have a terminal velocity
> KB> of 4,000 fpm, or about 48 mph. Of course, there is rolling and
> KB> aerodynamic drag, and there is prop drag too, so the engine
> KB> can only drag the plane along at, say, 30 mph, assuming a flat
> KB> smooth runway.
>
> KB> At 2,000 rpm, with no drag, the terminal velocity would be
> KB> 8,000 fpm, or about 85 mph.
>
> Hmmmm...so prop thrust is indeed only twice at double the
> rpm?...ideally speaking of course.
>
> The idealized (no viscosity etc.) math seems to say that it is linear,
> but intuitive feel says not.

Is aerodynamic drag of an airfoil linear with speed? Is airfoil lift
linear with speed? Since a prop is just an airfoil going in a circle,
why would you expect it to be linear?

Matt

[[email protected]]

>Kershner states that the maximum thrust force occurs when the plane is
>standing still (at a fixed throttle setting, I guess), and decreases as you
>go faster.

There are some fast homebuilts for which this won't be true.
They have fixed-pitch props with very high pitches, and the blade is
largely stalled at the start of the takeoff roll, making accelleration
dismal indeed. The pilots report that the airplane seems to come alive
at some point before liftoff when the prop blades finally get to work.
My own Jodel has an efficient wooden prop and I often note a small RPM
drop as the airplane accellerates through about 20 MPH. There's
something happening with the airflow through the blades, probably to do
with unstalling, or, perhaps (less likely) with leaving behind the
larger prop vortex generated in the static condition.

Dan

[Richard Lamb]

If you can find the engine performance plots you will see that the
percent of RPM and percent of power (HP or torque) are not at all
the same thing.

And it's torque that turns the propeller (not RPM).

1000 rpm might be near 1/2 RPM, but barely 10-20 percent max torque.

At full power (torque), the prop can deliver x number of pounds
thrust for any given airspeed. That's the most you'll get.

Rolling off RPM also rolls one down the torque curve.

And you are right, it's a very non-linear curve.


Richard

ps:

also on the torque curve, note that max torque and max HP are usually
NOT found at the same RPM...

ta

[Matt Whiting]

Richard Lamb wrote:

> If you can find the engine performance plots you will see that the
> percent of RPM and percent of power (HP or torque) are not at all
> the same thing.
>
> And it's torque that turns the propeller (not RPM).
>
> 1000 rpm might be near 1/2 RPM, but barely 10-20 percent max torque.
>
> At full power (torque), the prop can deliver x number of pounds
> thrust for any given airspeed. That's the most you'll get.
>
> Rolling off RPM also rolls one down the torque curve.
>
> And you are right, it's a very non-linear curve.
>
>
> Richard
>
> ps:
>
> also on the torque curve, note that max torque and max HP are usually
> NOT found at the same RPM...
>
> ta

It's been a while since I saw so many errors in so little text.

Matt

[Richard Lamb]

Matt Whiting wrote:
> Richard Lamb wrote:
>
>> If you can find the engine performance plots you will see that the
>> percent of RPM and percent of power (HP or torque) are not at all
>> the same thing.
>>
>> And it's torque that turns the propeller (not RPM).
>>
>> 1000 rpm might be near 1/2 RPM, but barely 10-20 percent max torque.
>>
>> At full power (torque), the prop can deliver x number of pounds
>> thrust for any given airspeed. That's the most you'll get.
>>
>> Rolling off RPM also rolls one down the torque curve.
>>
>> And you are right, it's a very non-linear curve.
>>
>>
>> Richard
>>
>> ps:
>>
>> also on the torque curve, note that max torque and max HP are usually
>> NOT found at the same RPM...
>>
>> ta
>
>
> It's been a while since I saw so many errors in so little text.
>
> Matt


That RPM and torque are NOT the same thing?

Or that at full power will give deliver full thrust?

Ot that the thrust delivered changes with airspeed?

Or that it's very non linear?

Very oversimplified, but go ahead and straighten me out, Matt.

Richard

[Matt Whiting]

Richard Lamb wrote:
> Matt Whiting wrote:
>
>> Richard Lamb wrote:
>>
>>> If you can find the engine performance plots you will see that the
>>> percent of RPM and percent of power (HP or torque) are not at all
>>> the same thing.
>>>
>>> And it's torque that turns the propeller (not RPM).
>>>
>>> 1000 rpm might be near 1/2 RPM, but barely 10-20 percent max torque.
>>>
>>> At full power (torque), the prop can deliver x number of pounds
>>> thrust for any given airspeed. That's the most you'll get.
>>>
>>> Rolling off RPM also rolls one down the torque curve.
>>>
>>> And you are right, it's a very non-linear curve.
>>>
>>>
>>> Richard
>>>
>>> ps:
>>>
>>> also on the torque curve, note that max torque and max HP are usually
>>> NOT found at the same RPM...
>>>
>>> ta
>>
>>
>>
>> It's been a while since I saw so many errors in so little text.
>>
>> Matt
>
>
>
> That RPM and torque are NOT the same thing?
>
> Or that at full power will give deliver full thrust?
>
> Ot that the thrust delivered changes with airspeed?
>
> Or that it's very non linear?
>
> Very oversimplified, but go ahead and straighten me out, Matt.
>
> Richard

Yes, horsepower and torque are absolutely not the same thing. The
following suggests that they are "At full power (torque)..."

Rolling off RPM may or may not roll you down the torque curve. If you
are running at an RPM above the torque peak, reducing RPM might actually
increase the torque available.

1000 RPM isn't 1/2 RPM. It may be close to 1/2 of the maximum allowable
RPM, which is what you hopefully intended to say.

Matt

[Tony]

Do you really think including factual data is likely to resolve the
question?

And do you know Dr Dan?

[darthpup]

Thrust is a direct relation of diameter x pitch x rpm.
Get a copy of H.Glauert's book: The elements of aerofoil and airscrew
theory.
isbn 052127494
Also, thrust is a function of the third power of the prop diameter so
changing the prop diameter on inch can have a major effect on thrust
and vice versa.

[Matt Whiting]

wrote:

>>Rolling off RPM may or may not roll you down the torque curve. If you
>>are running at an RPM above the torque peak, reducing RPM might actually
>>increase the torque available.
>
>
> Reducing RPM on a fixed-pitch prop will reduce the torque. Reducing
> RPM using the prop control on a constant-speed prop could increase
> torque as the RPM drops, depending on the engine's torque curve.

Not in all conditions. In SS level flight, yes.

> The torque curve is affected by volumetric efficiency, which is
> a bigger factor in high-RPM engines, less so in slow-turning aircraft
> engines. As RPM rises, the cylinder can't achieve anything near
> atmospheric pressure at the bottom of the intake stroke so that the
> amount of fuel/air mix is progressively reduced, and any horsepower
> increase with rising RPM is due to RPM only. The turbo or supercharger
> is the solution to the probem. In light airplanes, the turbo is more
> normally employed to alleviate altitude losses.

It's been a while since I studied engines in any detail, but I believe
there is more to it than just VE. I don't believe that bearing friction
is linear with RPM for example. Also, speed of the flame front becomes
and issue at higher RPM. I believe the drop-off in torque with RPM is a
function of a number of factors. VE is dominant, but not the only one.
Even turbocharged engines have a torque peak with a drop-off at some
point.

Matt


Matt

[[email protected]]

>I believe
>there is more to it than just VE. I don't believe that bearing friction
>is linear with RPM for example. Also, speed of the flame front becomes
>and issue at higher RPM. I believe the drop-off in torque with RPM is a
>function of a number of factors.

Yup you're right, there's more than just volumetric efficiency, but
flame front speed in these slow engines is still around 100 feet per
second, while average piston speed won't be much over 40 or 50 fps with
the midpoint travel being somewhat higher. The intake and exhaust
systems present more drag at higher RPMs and start to affect the
performance, and in many modern auto engines four valves per cylinder
are used to ease breathing.
I wonder if the new direct-drive diesel aircraft engines have much
higher torques in the right places?

Dan

[Morgans]

> wrote

> I wonder if the new direct-drive diesel aircraft engines have much
> higher torques in the right places?

Torque out the butt, and the torque stays high for a longer period of time
on the power stroke. In short, it will turn the same size prop of an engine
with nearly twice the HP.
--
Jim in NC

[Matt Whiting]

wrote:
>>I believe
>>there is more to it than just VE. I don't believe that bearing friction
>>is linear with RPM for example. Also, speed of the flame front becomes
>>and issue at higher RPM. I believe the drop-off in torque with RPM is a
>>function of a number of factors.
>
>
> Yup you're right, there's more than just volumetric efficiency, but
> flame front speed in these slow engines is still around 100 feet per
> second, while average piston speed won't be much over 40 or 50 fps with
> the midpoint travel being somewhat higher. The intake and exhaust
> systems present more drag at higher RPMs and start to affect the
> performance, and in many modern auto engines four valves per cylinder
> are used to ease breathing.
> I wonder if the new direct-drive diesel aircraft engines have much
> higher torques in the right places?
>
> Dan
>

It would seem they would. The high compression ratio gives most diesels
outstanding torque at fairly low RPMS, however, many also have a very
narrow torque peak.

Matt

[Brian Whatcott]

On Tue, 17 Jan 2006 22:48:09 -0800, "skyloon"
> wrote:


>Kershner states that the maximum thrust force occurs when the plane is
>standing still (at a fixed throttle setting, I guess), and decreases as you
>go faster. I do not understand this. Is it beacese AOA is largest? I am
>trying to see how this relates to power. Power would be force*distance/time
>or force*velocity. Maybe the thrust decreases slowly with airspeed, but the
>power still goes up as you go faster.
>
>This is just a hand waving argument. Please, anyone who knows more, feel
>free to correct this picture.
>
>Dave

I'll pick up on this one. There's a mechanics equation which is
specially straight-forward. It says if you apply a constant force to
an object,
and it moves in the direction of the force, then the work done is the
product of force times distance.

As expressed in the SI system, it's specially simple: F X D = W
gets the units of
F in Newtons times Distance in Meters equals work in joules
Even more interesting: F X V = P
force times speed = power.
In SI units again:
force in Newtons times speed in meters per second = power in Watts

OK that was the engineering/physics.

Now the application:
An airplane with a constant power recip prop engine.
Lets say the engine is putting out 90 HP say a C-152
90 HP = 90 X 746 watts = 67kW

Lets check the numbers at 10 mph, 50 mph and 100 mph
10 mph = 4.5 meters/sec
50 mph = 22.4 meters/sec
100 mph = 44.7 meters/sec

The unknown in the following equation is F
F X V = P or F = P/V

Now force is the same measure as thrust, so now we can
check available thust at these three speeds:

10 mph
F = 67000 W/4.5 M/Sec = 14890 Newtons
A newton, like a small apple weighs a quarter pound about.
So 14890 Newtons = 3340 pound That's a lot of thrust!

Now 50 mph
F = 67000/22.4 = 2990 Newtons or 671 lb.

Now 100 mph
F = 67000/44.7 = 1500 Newtons or 336 lb.

Or the general rule: the faster you go with constant power, the less
the thrust available.
Same applies to boats.

But think about planes with (some) jet engines,
these can be constant THRUST.

That means, the faster they go, the more HP they put out!
(A reason why jets on slow planes is not a great idea)

Brian Whatcott Altus OK

[David CL Francis]

On Sun, 22 Jan 2006 at 20:18:32 in message
>, Brian Whatcott
> wrote:

>But think about planes with (some) jet engines,
>these can be constant THRUST.
>
>That means, the faster they go, the more HP they put out!
>(A reason why jets on slow planes is not a great idea)

No, is not quite that easy. Another way of looking at force is that it
is rate of change of momentum.

In a simplified way the thrust of a jet engine comes from the change of
momentum from the air captured by the engine to the momentum of the air
that leaves the back of the engine. If you think of a Turbofan engine
then the fan is not so different from a propeller. The internal
efficiency of the two types of engine is somewhat different.

In any case the greatest propulsive efficiency comes from a momentum
change of a large mass of air with a very small velocity change.
--
David CL Francis

[Brian Whatcott]

On Tue, 24 Jan 2006 22:48:26 GMT, David CL Francis
> wrote:

>On Sun, 22 Jan 2006 at 20:18:32 in message
>, Brian Whatcott
> wrote:
>
>>But think about planes with (some) jet engines,
>>these can be constant THRUST.
>>
>>That means, the faster they go, the more HP they put out!
>>(A reason why jets on slow planes is not a great idea)
>
>No, is not quite that easy. Another way of looking at force is that it
>is rate of change of momentum.
>
>In a simplified way the thrust of a jet engine comes from the change of
>momentum from the air captured by the engine to the momentum of the air
>that leaves the back of the engine. If you think of a Turbofan engine
>then the fan is not so different from a propeller. The internal
>efficiency of the two types of engine is somewhat different.
>
>In any case the greatest propulsive efficiency comes from a momentum
>change of a large mass of air with a very small velocity change.


Hi David,

let's forget about jets and recips. Let's imagine a vehicle that is
provided with a constant thrust device.
Then, the faster it goes, the more hose power it provides.
You can take it to the bank

Brian

[David CL Francis]

On Thu, 26 Jan 2006 at 19:29:21 in message
>, Brian Whatcott
> wrote:

Hi Brian,

>Hi David,
>
>let's forget about jets and recips. Let's imagine a vehicle that is
>provided with a constant thrust device.
>Then, the faster it goes, the more hose power it provides.
>You can take it to the bank
>

No don't let us forget the basic ideas of propulsion.

A constant thrust device is doing one of two things;

1. Accelerating. In which case it is adding to its kinetic energy
and its power is going into that or

2. It reaches a constant speed against a constant drag and a steady
state occurs..

I presume you are not claiming that a constant thrust motor can generate
infinite power? In that case you would be right!

What do you have in mind as a constant thrust device? Newton's laws are
pretty good and I am not aware of any means of getting around them. They
only need adjustments at velocities and masses far beyond normal
terrestrial transport activities.

It is of course true that the efficiency of propulsion devices does vary
with speed and many other conditions.

--
David CL Francis

[Brian Whatcott]

On Mon, 30 Jan 2006 00:26:39 GMT, David CL Francis
> wrote:

[Brian]
>>let's forget about jets and recips. Let's imagine a vehicle that is
>>provided with a constant thrust device.
>>Then, the faster it goes, the more hose power it provides.
>>You can take it to the bank
>>
>
[David]
>No don't let us forget the basic ideas of propulsion.
>
>A constant thrust device is doing one of two things;
>
>1. Accelerating. In which case it is adding to its kinetic energy
> and its power is going into that or

...Good

>2. It reaches a constant speed against a constant drag and a steady
> state occurs..
>

....Good

>I presume you are not claiming that a constant thrust motor can generate
>infinite power? In that case you would be right!

I am claiming that engineers are familiar with two Newtonian equations
1) force times distance (in the direction of the force) = work
2) force times velocity in the direction of the force = power

I understand that it is non-intuitive to non-engineers that the
arrangement of
eqn 2) as
3) force = power / velocity is always true until large fractions of
c.

Though I cannot offer any further debate with you on this topic,
(unless you wish to pay me) it is helpful for you to know that thrust
is an equivalent term to force, and so for a vehicle with thrust
5 units and velocity 10 units, its power is 50 units.
5 = 50 / 10

Moreover, I am pretty sure you can work out the missing term in this
question
What is the power of a vehicle with thrust 5 units, and velocity 20
units? 5 = P / 20





(power = 100 units)

I urge you to contemplate the great simplicity of Newton's laws - and
their interesting practical applications.

Respectfully

Brian Whatcott Altus OK

[David CL Francis]

On Thu, 2 Feb 2006 at 17:59:33 in message
>, Brian Whatcott
> wrote:

>I am claiming that engineers are familiar with two Newtonian equations
>1) force times distance (in the direction of the force) = work
>2) force times velocity in the direction of the force = power
>

The above statements are true but not the whole story.

Are you considering a situation on a vehicle passing through an
atmosphere or a rocket in empty space? If there is no drag then the body
will accelerate as long as the thrust is present. Rockets can produce a
constant thrust but by their very nature the mass of the body reduces as
fuel is used up and the acceleration increases. What is the measure of
the power developed? The momentum change taking place as the fuel and
working fluids are expelled through the jet at their inherent velocity
or the distance the rocket moves? In space there is no resistance and
therefore the distance moved is irrelevant to power. The power is
transformed into a velocity change and kinetic energy. In an atmosphere
speed will increase to a steady state when thrust and drag are equal. Of
course they are both forces.

>I understand that it is non-intuitive to non-engineers that the
>arrangement of
>eqn 2) as
>3) force = power / velocity is always true until large fractions of
>c.
>
>Though I cannot offer any further debate with you on this topic,
>(unless you wish to pay me) it is helpful for you to know that thrust
>is an equivalent term to force, and so for a vehicle with thrust
>5 units and velocity 10 units, its power is 50 units.
>5 = 50 / 10

I do not feel inclined to continue this against such patronising
elementary statements.
--
David CL Francis